I'm self-studying Vershynin's High-dimensional probability book. I have a question about the proof of Theorem 8.1.4:
Let $(X_t)_{t\in T}$ be a mean-zero random process on a metric space $(T,d)$, with sub-gaussian increments. Then: $$ \mathbb{E}\sup_{t\in T}X_t\leq C K \sum_{k\in \mathbb{Z}}2^{-k}\sqrt{\log\mathcal{N}(T,d,2^{-k})} $$ Here, $C$ is a constant and $\mathcal{N}(T,d,2^{-k})$ the covering number of the set $T$ with balls of radius $2^{-k}$.
Question: In the proof, it is stated that without loss of generality, we may assume that $T$ is finite. Why can we do that?
By sub-gaussian increments, the following is meant:
Consider a random process $(x_t)_{t\in T}$ on a metric space $(T,d)$. We say that the process has sub-gaussian increments, if there exists a constant $K>0$, such that for all $t,x\in T$, $$ ||X_t-X_s||_{\psi_2}\leq Kd(t,s) $$ Here, $||\cdot||_{\psi_2}$ is the sub-gaussian norm.
I think I found an explanation: To avoid issues with measurability, Vershynin defines: $$ \mathbb{E}[\sup_{t\in T}X_t]:=\sup_{T'\in \mathcal{T}}\mathbb{E}[\sup_{t\in T'}X_t] $$ Where $\mathcal{T}$ is the set of finite subsets of $T$ (see footnote 3 on p.151). So, the supremum is only taken over finite sets $T'$. The proof in Theorem 8.1.4 for finite sets gives us: $$ \sup_{T'\in \mathcal{T}}\mathbb{E}[\sup_{t\in T'}X_t] \leq \sup_{T'\in \mathcal{T}}CK\sum_{k\in\mathbb{Z}}2^{-k}\sqrt{\log\mathcal{N}(T',d,2^{-k})} $$ And since $T'\subset T$, we have $\mathcal{N}(T',d,2^{-k})\leq \mathcal{N}(T,d,2^{-k})$, and so: $$ \leq CK\sum_{k\in\mathbb{Z}}2^{-k}\sqrt{\log\mathcal{N}(T,d,2^{-k})} $$
I am still curious whether this result holds more generally, i.e. without defining $\mathbb{E}[\sup_{t\in T}X_t]$ as Vershynin does. On the other hand, $\sup_{t\in T}X_t$ may be non-measurable, which would mean that the inequality does not make sense in the first place.